# Annotated Solution to the Jotto Puzzle of the Day

## May 6, 2003

guilt - 2
jimmy - 1
quirk - 2
broil - 2
vapid - 1
mazes - 0
novum - 1
cubic - 2

We've got one zero, so let's start with that. Taking out the letters in
**mazes** from the rest of the puzzle leaves us with:
guilt - 2
jiy - 1
quirk - 2
broil - 2
vpid - 1
novu - 1
cubic - 2

Not too much to work with, but we do have one word that lets us play with
only three possibilities (jiy=1), so let's start with that.
Since **i** is a lot more common than either **j** or **y**,
let's start by assuming the word has an **i**. Reducing the puzzle with
that in mind gives us:

i(gult=1)(jy=0)(qurk=1)(brol=1)(vpd=0)(novu=1)(cub=1)

One small thing to note is that we can disregard the second **c** in
**cubic** only because we are left with only one letter in that word.
If we were left with more than that, we would have to account for the
possibility that both **c**'s were in the word.

Reducing with the new zeros gives us:

i(gult=1)(qurk=1)(brol=1)(nou=1)(cub=1)

Since **u** is a common letter to a lot of these, let's try assuming
it's in there next. That's a nice techinque for clearing a lot of things
out. So, assuming there's a **u**, the puzzle reduces to:

iu(glt=0)(qrk=0)(brol=1)(no=0)(cb=0)

Notice that there can be no **b** (cb=0), **r** (qrk=0), **o**
(no=0), or **l** (glt=0). However, we need to have one of those letters
(brol=1), so we have a contradiction. This means that, if there is an
**i**, then there is not a **u**.

Let's try the **l** with the **i** next, since it's in a couple of
the remaining words. Remember that we know that there is no **u** if
there is an **i**, so we can reduce the puzzle to:

il(gt=0)(qrk=1)(bro=0)(no=1)(cb=1)

Reducing with the new zeros gives us:

il(qk=1)nc

It's a good bet that there's no **q** since there's no **u**. That
leaves us with **ilknc**. Playing with anagrams of that gives us the
answer: **clink**.

kevin@aq.org

Jotto Puzzle of the Day

Jotto

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Last modified: 07 May 2003